Fundamental domain of a group action #
A set s is said to be a fundamental domain of an action of a group G on a measurable space α
with respect to a measure μ if
-
sis a measurable set; -
the sets
g • sover allg : Gcover almost all points of the whole space; -
the sets
g • s, are pairwise a.e. disjoint, i.e.,μ (g₁ • s ∩ g₂ • s) = 0wheneverg₁ ≠ g₂; we require this forg₂ = 1in the definition, then deduce it for any twog₁ ≠ g₂.
In this file we prove that in case of a countable group G and a measure preserving action, any two
fundamental domains have the same measure, and for a G-invariant function, its integrals over any
two fundamental domains are equal to each other.
We also generate additive versions of all theorems in this file using the to_additive attribute.
- measurable_set : measurable_set s
- ae_covers : ∀ᵐ (x : α) ∂μ, ∃ (g : G), g +ᵥ x ∈ s
- ae_disjoint : ∀ (g : G), g ≠ 0 → ⇑μ ((g +ᵥ s) ∩ s) = 0
A measurable set s is a fundamental domain for an additive action of an additive group G
on a measurable space α with respect to a measure α if the sets g +ᵥ s, g : G, are pairwise
a.e. disjoint and cover the whole space.
- measurable_set : measurable_set s
- ae_covers : ∀ᵐ (x : α) ∂μ, ∃ (g : G), g • x ∈ s
- ae_disjoint : ∀ (g : G), g ≠ 1 → ⇑μ (g • s ∩ s) = 0
A measurable set s is a fundamental domain for an action of a group G on a measurable
space α with respect to a measure α if the sets g • s, g : G, are pairwise a.e. disjoint and
cover the whole space.
If for each x : α, exactly one of g • x, g : G, belongs to a measurable set s, then s
is a fundamental domain for the action of G on α.
If for each x : α, exactly one of g +ᵥ x, g : G, belongs to a measurable set
s, then s is a fundamental domain for the additive action of G on α.
If s and t are two fundamental domains of the same action, then their measures are equal.